What is the second derivative of f(x)=cos(e^-x)?

Feb 2, 2017

The second derivative is: f''(x) =e^-x((sin(e^-x)-cos(e^-x)).

Explanation:

To solve this problem we first need to remember the chain rule, which state that the derivative of $y = f \left(u\right)$ if $u = g \left(x\right)$ is
$y ' = f ' \left(u\right) \cdot u '$ or y'=(f'(g(x))*(g'(x))

In this case we have $y = \cos \left({e}^{-} x\right)$ which is $y = \cos \left(u\right)$ where $u = {e}^{-} x$

$f ' \left(x\right) = - \sin \left({e}^{-} x\right) \cdot \left(- {e}^{-} x\right)$ So:
$f ' \left(x\right) = {e}^{-} x \cdot \sin \left({e}^{-} x\right)$
For the second derivative, we use the product and chain rules:

$f ' ' \left(x\right) = - {e}^{-} x \cdot \sin \left({e}^{-} x\right) + {e}^{-} x \cdot \cos \left({e}^{-} x\right)$
Factoring, we have:
f''(x) =e^-x((sin(e^-x)-cos(e^-x))